Tưởng tượng nhé :v
(a+b)^2 + 3a+b)+10= t^2 +3t+10 ( đặ a+b=1)  = (t^2+3t+9/4) +31/4 >0  
=> Không thể phân tích :3

a²-b²+3a+3b=(a-b)(a+b)+3(a+b)=(a+b)(a-b+3)

=(a-b).(a+b)+3.(a+b)=(a+b).(a-b+3)

Ta có:A= 3a+3b-a^2-ab

=>A= (3a-a^2)+(3b-ab)

=>A= a(3-a)+b(3-a)

=>A= (a+b)(3-a)

tớ mới học lớp 6

mk mới hk lớp 6 thui ak

a) \(3a-3b+a^2-2ab+b^2\)

\(=3\left(a-b\right)+\left(a-b\right)^2\)

\(=\left(a-b\right)\left(a-b+3\right)\)

a)

3.(a-b) +2.(a-b ) =5 .(a-b )

câu b làm tương tự nha nhóm a^2 -2ab +b^2 vào 1nhoms và làm như câu a

mình làm nhầm 

a) \(\left(a-b\right).\left(a-b+3\right)\)

\(a^3-b^3+3a^2+3ab+b^2\)

\(=\left(a-b\right)\left(a^2+ab+b^2\right)+3\left(a^2+ab+b^2\right)\)

\(=\left(a-b+3\right)\left(a^2+ab+b^2\right)\)

\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)

1/Tự chép lại đb nha :v

 =a² - 9b²+2ab+3a²-8b²-12ab+6ab-3b²-2a²+ab

= 2a²-3ab-20b²

= (2a²+5ab) - (8ab+20b²)

= a(2a+5b) - 4b(2a+5b)

=(2a+5b)(a-4b)

câu 2 tương tự nhé :)

a/ 3(a + b) + c(a + b) = (a + b)(3 + c)

b/ (a - b)² - c² = (a - b - c)(a - b + c)